36 terms

How many grams are there in 3.58 short tons? Show all work, all units, 1lb= 16 oz, 1 short ton= 2000lb.

3.58 short ton= 2000 lb/ 1 short ton X 16 oz./ 1lb X 1 g/ 0.03527 oz = 3.25E 6 grams

Chap. 1

Chap. 1

Sample A- Mass of Sample:1.518g MofFE:1.094g MofO: 0.424g

Sample B- 2.056, 1.449g, 0.607g

Sample c- 1.873g, 1.335g, 0.538g

Sample B- 2.056, 1.449g, 0.607g

Sample c- 1.873g, 1.335g, 0.538g

Sample A- 1.094/ 0.424= 2.58

Sample B- 1.449/ 0.607= 2.38

Sample C- 1.335/ 0.538= 2.48

**No because the ratios of M(iron) to M(oxygen) are all different.**

Chap. 1

Sample B- 1.449/ 0.607= 2.38

Sample C- 1.335/ 0.538= 2.48

Chap. 1

The density of osmium metal ( a platinum- group metal) is 22.5 g/cm^3. Express the density in SI units (kg/m^3).

22.5g/cm^3 = 10^-3 kg/1g x 10^6 cm^3/ 1m^3= 2.25x10^4 kg/m^3

Chap. 1

Chap. 1

Calculate the relative abundance of each isotope of Boron from the following data:

Isotope Mass: Isotope Mass (amu): Average Atomic

B-10: 10.013=

------------------------------ 10.812

B-11: 11.009=

Isotope Mass: Isotope Mass (amu): Average Atomic

B-10: 10.013=

------------------------------ 10.812

B-11: 11.009=

B-10= x

B-11= y

*x+y= 1--> x= 1-y

** x(10.013)+y(11.009)= 10.812

----> (1-y)(10.013)+ 11.009y= 10.812

-----> 10.013- 10.013y+11.009y = 10.812

-10.013 -10.013

----> .996y=.799

***-----> y=.8022088353X100= 80.221% (B-11)

**** B-10= 100%- 80.221%= 19.780%

Chap. 1

B-11= y

*x+y= 1--> x= 1-y

** x(10.013)+y(11.009)= 10.812

----> (1-y)(10.013)+ 11.009y= 10.812

-----> 10.013- 10.013y+11.009y = 10.812

-10.013 -10.013

----> .996y=.799

***-----> y=.8022088353X100= 80.221% (B-11)

**** B-10= 100%- 80.221%= 19.780%

Chap. 1

An element has three naturally occurring isotopes with the following masses and abundances:

Isotopic Mass (amu)----- Fractional Abundance

I- 38.964------ 0.9326

II- 39.964------ 1.000X 10^-4

III- 40.962------ 0.0673

Isotopic Mass (amu)----- Fractional Abundance

I- 38.964------ 0.9326

II- 39.964------ 1.000X 10^-4

III- 40.962------ 0.0673

(Isotope Mass)(Fractional Abundance)=

I- 38.964(0.9326) = 36.3378264

II- 39.964(1.000X 10^-4)= .0039964

III- 40.962(0.0673)= 2.7567426

----> I + II + III= 39.0985654

**------> 39.10 Potassium

Chap. 1

I- 38.964(0.9326) = 36.3378264

II- 39.964(1.000X 10^-4)= .0039964

III- 40.962(0.0673)= 2.7567426

----> I + II + III= 39.0985654

**------> 39.10 Potassium

Chap. 1

A 1.50 g sample of nitrous oxide (an anesthetic, sometimes called laughing gas) contains 2.05 X 10^22 N2O molecules. How many nitrogen atoms are in this sample? How many nitrogen atoms are in 1.00 g of nitrous oxide?

NO= 1.50g

N2O= 2.05 x 10^22

*---> 1.50g NO/ 2.05X10^22= 1g/ x

------> 1.3666667 X 10^22 nitrogen atoms--> 1.37 X 10^22

= ------> 2(1.37x10^22) = 2.74x 10^22 nitrogen atoms in 1 g of N2O

** Nitrogen=

----> 2.05x10^22 N2O= 2N atoms/1 N2O= 4.10x10^22 nitrogen atoms

Chap. 1

N2O= 2.05 x 10^22

*---> 1.50g NO/ 2.05X10^22= 1g/ x

------> 1.3666667 X 10^22 nitrogen atoms--> 1.37 X 10^22

= ------> 2(1.37x10^22) = 2.74x 10^22 nitrogen atoms in 1 g of N2O

** Nitrogen=

----> 2.05x10^22 N2O= 2N atoms/1 N2O= 4.10x10^22 nitrogen atoms

Chap. 1

How many significant figures are there in the value 4,750,330?

a. 7

b. 6

c. 5

d. 4

e. 3

a. 7

b. 6

c. 5

d. 4

e. 3

b. 6

Chap. 1

Chap. 1

Which of the following sets of numbers have the same number of sig figs?

1. 3.55E-3 2. 3.55E3 3. 3,550,000

a. 1 and 2

b. 1 and 3

c. 2 and 3

d, 1,2,3

e. none of them

1. 3.55E-3 2. 3.55E3 3. 3,550,000

a. 1 and 2

b. 1 and 3

c. 2 and 3

d, 1,2,3

e. none of them

d. 1,2,3

Chap. 1

Chap. 1

The number of significant figures that should be reported for the answer the mathematical computation 142.000- 41.9903 is

a. 2

b. 3

c. 4

d. 5

e. 6

a. 2

b. 3

c. 4

d. 5

e. 6

d. 5

Chap. 1

Chap. 1

Liquid propane boils at -42C. What is its boiling point on the Kelvin scale?

a. 231k

b. 256k

c. 273k

d. 315k

e. 345k

a. 231k

b. 256k

c. 273k

d. 315k

e. 345k

a. 231k

Chap. 1

Chap. 1

The average velocity of oxygen molecule at 1000 is 8X 10^4 cm/s. Which of the following calculations would convert this value to the velocity in miles per hour?

e. 8X10^4 cm/s X 1 in/ 2.54 cm X 1 ft/ 12 in X 1 mi/ 5280 ft X 3600s/ 1 hr

Chap. 1

Chap. 1

A piece of indium weighing 15.442g is placed in 49.7 cm^3 of ethyl alcohol (d= 0.789 g/ cm^3) in a graduated cylinder. The alcohol level increases to 51.8 cm^3. The best value for density of indium from these data:

a. 7.353 g/ cm^3

b. 7.35 g/cm^3

c. 7.4 g/ cm^3

d. 9.21 g/ cm^3

e. 9.2 g/ cm^3

a. 7.353 g/ cm^3

b. 7.35 g/cm^3

c. 7.4 g/ cm^3

d. 9.21 g/ cm^3

e. 9.2 g/ cm^3

c. 7.4 g/cm^3

Chap. 1

Chap. 1

How many protons, neutrons, and electrons are in the tin(II) ion 119, 50 Sn^ 2+? (p,n,e)

a. 119, 50, 119

b. 50, 69, 50

c. 50, 69, 48

d. 69, 50, 69

e. 50, 119, 52

a. 119, 50, 119

b. 50, 69, 50

c. 50, 69, 48

d. 69, 50, 69

e. 50, 119, 52

c. 50p, 69n, 48e

Chap. 1

Chap. 1

A certain element is listed as having 63.5 atomic mass units. It is probably true that it:

a. a mixture of isomers

b. a mixture of allotropes

c. a mixture of neutrons

d. a mixture of ions

e. a mixture of isotopes

a. a mixture of isomers

b. a mixture of allotropes

c. a mixture of neutrons

d. a mixture of ions

e. a mixture of isotopes

e. a mixture of isotopes

Chap. 1

Chap. 1

The mass spectrum of an element with two naturally occurring isotopes is shown below. Its atomic mass would be best estimated as (chart).

c. 63.6

Chap. 1

Chap. 1

Choose the pair of names and formulas that DO NOT match:

a. Sodium Sulfite Na2S

b. Calcium fluoride CaF2

c. Potassium permanganate KMnO4

d. Aluminum bromide AlBr3

e. Iron (III) oxide Fe2O3

a. Sodium Sulfite Na2S

b. Calcium fluoride CaF2

c. Potassium permanganate KMnO4

d. Aluminum bromide AlBr3

e. Iron (III) oxide Fe2O3

a. Sodium Sulfite--- Na2S

Chap. 1

Chap. 1

The formula neodymium sulfate is Nd2(SO4). On the basis of this information, the formula for the nitride of neodymium would be expected to be:

a. Nd2(NO2)3

b. Nd2N2

c. Nd(NO3)3

d. Nd(NO2)3

e. NdN

a. Nd2(NO2)3

b. Nd2N2

c. Nd(NO3)3

d. Nd(NO2)3

e. NdN

e. NdN

Chap. 1

Chap. 1

Treatment of sodium borohydrate with sulfuric acid is a convenient method for the preparation of diborane:

__ NaBH4+ ___H2SO4---. ___ B2H6+ H2+ Na2SO4

a. 1

b. 2

c. 3

d. 4

e. 5

__ NaBH4+ ___H2SO4---. ___ B2H6+ H2+ Na2SO4

a. 1

b. 2

c. 3

d. 4

e. 5

b.2

Which of the following equations is (are) balanced?

1. NaCl + Pb(NO3)2---> PbCl2 + NaNO3

2. (NH4)2Cr2O7 ---> N2 + 4H2O + Cr2O3

3. FeO4 +3CO ---> 3Fe + 3 Co2

a. 1

b. 2

c. 3

d. 1 and 2

e. 1,2, 3

1. NaCl + Pb(NO3)2---> PbCl2 + NaNO3

2. (NH4)2Cr2O7 ---> N2 + 4H2O + Cr2O3

3. FeO4 +3CO ---> 3Fe + 3 Co2

a. 1

b. 2

c. 3

d. 1 and 2

e. 1,2, 3

b. 2

What is the ratio of oxygen atoms to hydrogen atoms in the mineral carnotite K2(UO2)3 (VO4)2 X 3H2O:

a. 8:3

b. 8:6

c. 9:6

d. 17:3

e. 17:6

a. 8:3

b. 8:6

c. 9:6

d. 17:3

e. 17:6

e. 17:6

**END CHAP.1**

Calculate the percentage composition for each of the following componds (to three sig digs):

(a) CO2

(a) CO2

*C+O2= 12.01 + 2(15.99) = 44.008

** C- 12.01/ 44.008= .273 X 100= 27.3%

*** O2 - 2(15.99)/ 44.008 = .727 X 100= 72.7%

** C- 12.01/ 44.008= .273 X 100= 27.3%

*** O2 - 2(15.99)/ 44.008 = .727 X 100= 72.7%

Malonic acid is used in the manufacture of barbiturates (sleeping pills). The composition of the acis is 34.6% C, 3.9% H, and 61.5% O. What is Malonic acid's empirical formula?

*C- 34.6g x 1mol/ 12.01g C = 2.88

H- 3.9g H X 1mol/ 1.01g H= 3.86

O- 61.5 O x 1 mol/ 15.999g O = 3.84

** C- 2.88/ 2.88= 1 X 3 = 3

H- 3.68/2.88= 1.34 X 3= 4.02--> 4

O- 3.84/ 2.88= 1.33 X 3 = 4

**** C3H4O4 ******

H- 3.9g H X 1mol/ 1.01g H= 3.86

O- 61.5 O x 1 mol/ 15.999g O = 3.84

** C- 2.88/ 2.88= 1 X 3 = 3

H- 3.68/2.88= 1.34 X 3= 4.02--> 4

O- 3.84/ 2.88= 1.33 X 3 = 4

**

Nickel (II) cholride reacts with sodium phosphate to precipitate nickel (II) phosphate and another.

(a) Write a balanced equation for this reaction.

(b) What is the amount (in grams) of nickel (II) phosphate produced from 42.6g of sodium phosphate?

(a) Write a balanced equation for this reaction.

(b) What is the amount (in grams) of nickel (II) phosphate produced from 42.6g of sodium phosphate?

(a) 3NiCl2 + 2Na3PO4 ---> Ni3(PO4)2 + 6NaCl

(b) 42.6g Na3PO4 x 1mol/ 163.9392 Na3PO4 X 1mol/ Ni3(PO4)2/ 2 mol Na3PO4 X 366g Na3PO4/ 1mol Na3(PO4)2

= 43.5 g Ni3(PO4)2

(b) 42.6g Na3PO4 x 1mol/ 163.9392 Na3PO4 X 1mol/ Ni3(PO4)2/ 2 mol Na3PO4 X 366g Na3PO4/ 1mol Na3(PO4)2

= 43.5 g Ni3(PO4)2

Nitric acid HNO3 is manufatured by the Ostwald process in which nitrogen dioxide, NO2, reacts with water.

3NO2 (g) + H2O ---> 2HNO3 + NO

Suppose a vessel contains 4.05 grams NO2 and 6.72 g H20.

(a) What is the limiting reactant?

(b) How many grams of HNO3 could be obtained?

3NO2 (g) + H2O ---> 2HNO3 + NO

Suppose a vessel contains 4.05 grams NO2 and 6.72 g H20.

(a) What is the limiting reactant?

(b) How many grams of HNO3 could be obtained?

(a) NO2

(b)

NO2-- 4.05g NO2 X 1 mol/ 46.0047g NO2 x 2 mol HNO3/ 3 mol NO2= .05868 = .059

H2O-- 6.72g H2O X 1 mol H2O/ 18.01g H2O X 2 mol HNO3/ 1 mol HNO3 = .07458= .075

Limiting reactant: .059 mol HNO3 X 63.0137 g HNO3/ 1 mol HNO3= 3.7178= 3.72g HNO3

(b)

NO2-- 4.05g NO2 X 1 mol/ 46.0047g NO2 x 2 mol HNO3/ 3 mol NO2= .05868 = .059

H2O-- 6.72g H2O X 1 mol H2O/ 18.01g H2O X 2 mol HNO3/ 1 mol HNO3 = .07458= .075

Limiting reactant: .059 mol HNO3 X 63.0137 g HNO3/ 1 mol HNO3= 3.7178= 3.72g HNO3

Write the oxidation- reduction half reactions for the following reactions and balance the overall reaction.

(a) Mn^2+ + BiO3^- ----> MnO4^- +Bi^3+

(b) Cr2O7^-2 + I^- ---> Cr^3+ + IO3^-

(a) Mn^2+ + BiO3^- ----> MnO4^- +Bi^3+

(b) Cr2O7^-2 + I^- ---> Cr^3+ + IO3^-

(a) [ Mn^ 2+ ===> Mn^7+ + 5e- ]= 2Mn ---> 2MnO4 + 10 e-

---> 5[ 2e- +BiO3^- ===> Bi^3+] = 70 e- + 5BiO3---> 5Bi

*---> 2Mn + 5BiO3 ---> 5Bi + 2MnO4

(b) [ 6e- + Cr2O7 ---> 2Cr^3+] ===> 6e-+ Cr2)7----> Cr^3+

-----> [I^- (+6) ---> IO3^-] 9 ===> 9I---> 9IO3 + 9e-

**----> 9I +Cr2O7---> Cr+ 9IO3

---> 5[ 2e- +BiO3^- ===> Bi^3+] = 70 e- + 5BiO3---> 5Bi

*---> 2Mn + 5BiO3 ---> 5Bi + 2MnO4

(b) [ 6e- + Cr2O7 ---> 2Cr^3+] ===> 6e-+ Cr2)7----> Cr^3+

-----> [I^- (+6) ---> IO3^-] 9 ===> 9I---> 9IO3 + 9e-

**----> 9I +Cr2O7---> Cr+ 9IO3

What is the percentage of chlorine in DDT, C14H9Cl5?

d. 50.0

A given hydrocarbon is converted completely to water and carbon dioxide, and the mole ratio of H2O to CO2 is 1.33:1.00. The Hydrocarbon could be?

e. C3H8

Which of the following contains the greatest mass of chromium?

b. 25.0g CrO2

Ammonia can be made by reaction of water with calcium cyanamide:

CaCN2 + 3H2O ---> CaCO3 + 2NH3

When the equation is properly balanced, the sum of the coefficients is:

CaCN2 + 3H2O ---> CaCO3 + 2NH3

When the equation is properly balanced, the sum of the coefficients is:

d.7

All of the following are weak acids except:

e. HNO3

All of the following are strong acids in aqueous solution except:

a. HClO2

All the following are strong electrolytes except:

b. H3PO4

The oxidation number of chromium in sodium chromite, NaCrO2, is

e. +3

Which of the following reactions is an oxidation-reduction reaction?

b. NH4NO3 ---> N2O + 2H2O

The formation of an anion from an atom would be called:

e. reduction

**END of 3&4**

SnO2 + 2CO --> Sn + 2CO2 DeltaH= +14.7Kj

a. Calculate the amount of heat absorbed when 5.60g of tin are formed.

b. Calculate the volume of CO2 evolved (measured at 745mmHg and 23C) when 5.60 g of tin are formed.

a. Calculate the amount of heat absorbed when 5.60g of tin are formed.

b. Calculate the volume of CO2 evolved (measured at 745mmHg and 23C) when 5.60 g of tin are formed.

a. 5.60g Sn x 1mol Sn/ 118.71g Sn x 2mol CO2/ 1 mol Sn= .094 mol

PV= nRT ---> V= nRT/P

===> .094 mol (0.0821)(296k)/ (745/760) = 2.2843504/ .98026= 2.33 g/L

PV= nRT ---> V= nRT/P

===> .094 mol (0.0821)(296k)/ (745/760) = 2.2843504/ .98026= 2.33 g/L